Source: High School Physics Olympiad
Problem Sets:
Problem
A particle moves in the $Oxy$ plane with motion equations $x = 3t$, $y = 8 - t^2$, where $x$ and $y$ are in meters and $t$ is in seconds.
- Write the position vector of the particle at time $t$ and find the trajectory of the particle.
- Calculate the average velocity during the second second (from $t = 1$ s to $t = 2$ s).
- Calculate the instantaneous velocity and instantaneous acceleration at the end of the 1st second and the end of the 2nd second. At what time does the particle have minimum speed?
- At what time $t$ is the particle closest to the origin? Calculate this distance.
Q1: The position vector is $\vec{r} = 3t\hat{i} + (8 - t^2)\hat{j}$ m. Eliminating $t = x/3$ gives the trajectory $y = 8 - \dfrac{x^2}{9}$, a parabola.
Q2: $\vec{r}(1) = (3, 7)$ m and $\vec{r}(2) = (6, 4)$ m, so $\Delta\vec{r} = (3, -3)$ m over $\Delta t = 1$ s:
$$\bar{v} = 3\hat{i} - 3\hat{j} \text{ m/s}, \qquad |\bar{v}| = 3\sqrt{2} \approx 4.2 \text{ m/s}$$Q3: The $x$-motion is uniform with $v_x = 3$ m/s; the $y$-motion is uniformly accelerated with $a_y = -2$ m/s$^2$ and $v_y = -2t$. So $\vec{v} = 3\hat{i} - 2t\hat{j}$ m/s and $\vec{a} = -2\hat{j}$ m/s$^2$ (constant).
At $t = 1$ s: $\vec{v} = 3\hat{i} - 2\hat{j}$ m/s, $|\vec{v}| = \sqrt{13} \approx 3.6$ m/s. At $t = 2$ s: $\vec{v} = 3\hat{i} - 4\hat{j}$ m/s, $|\vec{v}| = 5$ m/s.
Since $|\vec{v}| = \sqrt{9 + 4t^2}$ grows with $t$, the speed is minimum at $t = 0$: $v_{min} = 3$ m/s.
Q4: The squared distance is $r^2 = (3t)^2 + (8 - t^2)^2 = t^4 - 7t^2 + 64$. With $u = t^2$ this is the quadratic $u^2 - 7u + 64$, minimized at $u = \frac{7}{2}$, so:
$$t = \sqrt{7/2} \approx 1.87 \text{ s}, \qquad r_{min} = \sqrt{(7/2)^2 - 7(7/2) + 64} = \sqrt{51.75} \approx 7.2 \text{ m}$$